10
MAXIMUM AND MINIMUM
VALUES
The turning points of a graph
WE SAY THAT A FUNCTION f(x) has a relative maximum value at x = a,
if f(a) is greater than any value immediately preceding or follwing.
We call it a "relative" maximum because other values of the function may in fact be greater.
We say that a function f(x) has a relative minimum value at x = b,
if f(b) is less than any value immediately preceding or follwing.
Again, other values of the function may in fact be less. With that understanding, then, we will drop the term relative.
The value of the function, the value of y, at either a maximum or a minimum is called an extreme value.
Now, what characterizes the graph at an extreme value?
The tangent to the curve is horizontal. We see this at the points A and B. The slope of each tangent line -- the derivative when evaluated at a or b -- is 0.
f '(x) = 0.
Moreover, at points immediately to the left of a maximum -- at a point C -- the slope of the tangent is positive: f '(x) > 0. While at points immediately to the right -- at a point D -- the slope is negative: f '(x) < 0.
In other words, at a maximum, f '(x) changes sign from + to − .
At a minimum, f '(x) changes sign from − to + . We can see that at the points E and F.
We can also observe that at a maximum, at A, the graph is concave downward. (Topic 14 of Precalculus.) While at a minimum, at B, it is concave upward.
A value of x at which the function has either a maximum or a minimum is called a critical value. In the figure --
-- the critical values are x = a and x = b.
The critical values determine turning points, at which the tangent is parallel to the x-axis. The critical values -- if any -- will be the solutions to f '(x) = 0.
Example 1. Let f(x) = x2 − 6x + 5.
Are there any critical values -- any turning points? If so, do they determine a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum?
Solution. f '(x) = 2x − 6 = 0 implies x = 3. (Lesson 9 of Algebra.)
x = 3 is the only critical value. It is the x-coördinate of the turning point. To determine the y-coördinate, evaluate f at that critical value -- evaluate f(3):
| f(x) | = | x2 − 6x + 5 |
| f(3) | = | 32 − 6· 3 + 5 |
| = | −4. | |
The extreme value is −4. To see whether it is a maximum or a minimum, in this case we can simply look at the graph.
f(x) is a parabola, and we can see that the turning point is a minimum.
By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, −4).
But we will not always be able to look at the graph. The algebraic condition for a minimum is that f '(x) changes sign from − to + . We see this at the points E, B, F above. The value of the slope is increasing.
Now to say that the slope is increasing, is to say that, at a critical value, the second derivative (Lesson 9) -- which is rate of change of the slope -- is positive.
Again, here is f(x):
| f(x) | = | x2 − 6x + 5. |
| f '(x) | = | 2x − 6. |
| f ''(x) | = | 2. |
f '' evaluated at the critical value 3 -- f''(3) = 2 -- is positive. This tells us algebraically that the critical value 3 determines a minimum.
Sufficient conditions
We can now state these sufficient conditions for extreme values of a function at a critical value a:
The function has a minimum value at x = a if f '(a) = 0
and f ''(a) = a positive number.
The function has a maximum value at x = a if f '(a) = 0
and f ''(a) = a negative number.
In the case of the maximum, the slope of the tangent is decreasing -- it is going from positive to negative. We can see that at the points C, A, D.
Example 2. Let f(x) = 2x3− 9x2 + 12x − 3.
Are there any extreme values? First, are there any critical values -- solutions to f '(x) = 0 -- and do they determine a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum? Where are the turning points?
| Solution. f '(x) = 6x2 − 18x + 12 | = | 6(x2 − 3x + 2) |
| = | 6(x − 1)(x − 2) | |
| = | 0 | |
implies:
x = 1 or x = 2.
Those are the critical values. Does each one determine a maximum or does it determine a minimum? To answer, we must evaluate the second derivative at each value.
| f '(x) | = | 6x2 − 18x + 12. |
| f ''(x) | = | 12x − 18. |
| f ''(1) | = | 12 − 18 = −6. |
The second derivative is negative. The function therefore has a maximum at x = 1.
To find the y-coördinate -- the extreme value -- at that maximum we evaluate f(1):
| f(x) | = | 2x3− 9x2 + 12x − 3 |
| f(1) | = | 2 − 9 + 12 − 3 |
| = | 2. | |
The maximum occurs at the point (1, 2).
Next, does x = 2 determine a maximum or a minimum?
| f ''(x) | = | 12x − 18. |
| f ''(2) | = | 24 − 18 = 6. |
The second derivative is positive. The function therefore has a minimum at x = 2.
To find the y-coördinate -- the extreme value -- at that minimum, we evaluate f(2):
| f(x) | = | 2x3 − 9x2 + 12x − 3. |
| f(2) | = | 16 − 36 + 24 − 3 |
| = | 1. | |
The minimum occurs at the point (2, 1).
Here in fact is the graph of f(x):
Solutions to f ''(x) = 0 indicate a point of inflection at those solutions, not a maximum or minimum. An example is y = x3. y'' = 6x = 0 implies x = 0. But x = 0 is a point of inflection in the graph of y = x3, not a maximum or minimum.
Another example is y = sin x. The solutions to y'' = 0 are the multiplies of π, which are points of inflection.
Problem 1. Find the coördinates of the vertex of the parabola,
y = x2 − 8x + 1.
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!
y' = 2x − 8 = 0.
That implies x = 4. That's the x-coördinate of the vertex. To find the y-coördinate, evaluate y at x = 4:
y = 42 − 8· 4 + 1 = −15.
The vertex is at (4, −15).
Problem 2. Examine each function for maxima and minima.
a) y = x3 − 3x2 + 2.
y' = 3x2 − 6x = 3x(x − 2) = 0 implies
x = 0 or x = 2.
y''(x) = 6x − 6.
y''(0) = −6.
The second derivative is negative. That means there is a maximum at x = 0. That maximum value is
y(0) = 2.
Next,
y''(2) = 12 − 6 = 6.
The second derivative is positive. That means there is a minimum at x = 2. That minimum value is
y(2) = 23 − 3· 22 + 2 = 8 − 12 + 2 = −2.
b) y = −2x3 − 3x2 + 12 x + 10.
At x = 1 there is a maximum of y = 17.
At x = −2 there is a minimum of y = −10.
c) y = 2x3 + 3x2 + 12 x − 4.
Since f '(x) = 0 has no real solutions, there are no extreme values.
d) y = 3x4− 4x3 − 12x2 + 2.
At x = 0 there is a maximum of y = 2.
At x = −1 there is a minimum of y = −3.
At x = 2 there is a minimum of y = −30.
Next Lesson: Applications of maximum and minimum values
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